LESSON: Doppler Effect


Has this ever happened to you? You hear a siren from a few blocks away. The source is a police car that is racing in your direction. As the car approaches, zooms past you, and then speeds off into the distance, the sound of its siren keeps changing in pitch. First the siren gets higher in pitch, and then it suddenly gets lower. Do you know why this happens? The answer is the Doppler effect.

What Is the Doppler Effect?

The Doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener. (It can also occur when the sound source is stationary and the listener is moving.) The diagram below shows how the Doppler effect occurs. The sound waves from the police car siren travel outward in all directions. Because the car is racing forward (to the left), the sound waves get bunched up in front of the car and spread out behind it. Sound waves that are closer together have a higher frequency, and sound waves that are farther apart have a lower frequency. The frequency of sound waves, in turn, determines the pitch of the sound. Sound waves with a higher frequency produce sound with a higher pitch, and sound waves with a lower frequency produce sound with a lower pitch.


Experiencing the Doppler Effect

As the car approaches listener A, the sound waves get closer together, increasing their frequency. This listener hears the pitch of the siren get higher. As the car speeds away from listener B, the sound waves get farther apart, decreasing their frequency. This listener hears the pitch of the siren get lower. You can experience the Doppler effect with a moving siren in the following animation:

Q: What will the siren sound like to listener A after the police car passes him?

A: The siren will suddenly get lower in pitch because the sound waves will be much more spread out and have a lower frequency.

Summary

  • The Doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener.
  • As the source of sound waves approaches a listener, the sound waves get closer together, increasing their frequency and the pitch of the sound. The opposite happens when the source of sound waves moves away from the listener.

Vocabulary

  • Doppler effect: Change in the frequency and pitch of sound that occurs when the source of the sound is moving relative to the listener.



Key Equations

Doppler Shifts:f_o &= f \frac{v+v_o}{v-v_s} && f_o \mathrm{~(observed~frequency)~is~shifted~up~when~source~and~observer~moving~closer}\\f_o &= f \frac{v-v_o}{v+v_s} && f_o \mathrm{~(observed~frequency)~is~shifted~down~when~source~and~observer~moving~apart, where}\\& && v \text{ is the speed of sound, } v_s \text{ is the speed of the source, and } v_o \text{ is the speed of the observer}

Guidance
The Doppler Effect occurs when either the source of a wave or the observer of a wave (or both) are moving. When a source of a wave is moving towards you, the apparent frequency of the wave you detect is higher than that emitted. For instance, if a car approaches you while playing a note at 500 Hz, the sound you hear will be slightly higher frequency. The opposite occurs (the frequency observed is lower than emitted) for a receding wave or if the observer moves away from the source. It’s important to note that the speed of the wave does not change –it’s traveling through the same medium so the speed is the same. Due to the relative motion between the source and the observer the frequency changes, but not the speed of the wave. Note that while the effect is similar for light and electromagnetic waves the formulas are not exactly the same as for sound.

Example 1

Question: How fast would a student playing an A note (440\mathrm{Hz}) have to move towards you in order for you to hear a G note (784\mathrm{Hz})?

Answer: We will use the Doppler shift equation for when the objects are getting closer together and solve for the speed of the student (the source).

f_o=f(\frac{v+v_o}{v-v_s}) \Rightarrow f_o\times (v-v_s)=f\times (v+v_o) \Rightarrow vf_o-v_sf_o=f\times (v+v_o) \Rightarrow v_s=-(\frac{f\times (v+v_o)-vf_o}{f_o})

Now we plug in the known values to solve for the velocity of the student.

v_s=-(\frac{f\times (v+v_o)-vf_o}{f_o})=-(\frac{440\mathrm{Hz}\times (343\mathrm{m/s}+0\mathrm{m/s})-343\mathrm{m/s}\times 784\mathrm{Hz}}{784\mathrm{Hz}})=151\mathrm{m/s}


Watch the following videos for more examples of calculations:


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Last modified: Friday, 3 March 2017, 12:21 PM